Writing: $$f\stackrel{(g_1,g_2)}{\to}f'\tag1$$ where $f,f'$ are objects of arrow category $\mathcal C^{\to}$ and pair $(g_1,g_2)$ is an element of homset $\mathcal C^{\to}(f,f')$ represents a commuting diagram pictured in your question.
We have the functor $\mathbf{dom}:\mathcal C^{\to}\to\mathcal C$ prescribed by:$$[f\stackrel{(g_1,g_2)}{\to}f']\mapsto[\mathsf{dom}f\stackrel{g_1}{\to}\mathsf{dom}f']$$
And we have the functor $\mathbf{cod}:\mathcal C^{\to}\to\mathcal C$ prescribed by:$$[f\stackrel{(g_1,g_2)}{\to}f']\mapsto[\mathsf{cod}f\stackrel{g_2}{\to}\mathsf{cod}f']$$
In order to prove that $\mathbf{dom}$ and $\mathbf{cod}$ are functors it must be shown both of them respect identities and composition.
If $(1)$ stands for an identity then $f=f'$ and $g_1,g_2$ are both identities in $\mathcal C$. This guarantees that identities are respected.
By composition we must expand $(1)$ to: $$f\stackrel{(g_1,g_2)}{\to}f'\text{ and }f'\stackrel{(g'_1,g'_2)}{\to}f''\tag2$$with commuting squares.
Then we have: $$(g'_1,g'_2)\circ(g_1,g_2)=(g'_1\circ g_1,g'_2\circ g_2)$$assuring that composition is respected.
There are a least a couple problems with your construction.
First, your construction uses $U$ and the objects $a$, $b$ and $z$. Where do they come from? If $U$ is still a functor and $a$, $b$ and $z$ are objects of $\mathcal{D}$, then we're back where we started with regards to the original definition. You may just be treating $U(a)$ and the rest as formal symbols, which is fine, but that means that this has nothing to do with the original construction.
Second, you don't have a complete list of the objects of the arrow category of $\mathcal{C}$. You said we morphisms $U(g) : U(a) \to U(b)$ and $U(h) : U(a) \to U(z)$, which means there are (at least) more objects in the arrow category. Also, you've neglected to mention the four identity arrows for $x$, $U(a)$, $U(b)$ and $U(z)$.
Moreover, there are a few compositions missing, though you may be implicitly saying that $U(g) \circ f_3 = f_2$ and $U(h) \circ f_3 = f_1$. We have to assume that this is true for $U(g)$ (really $\langle id_x, U(g) \rangle$) and $U(h)$ to be morphisms in the arrow category.
Now including the missing objects, $\langle x, f_3, U(a) \rangle$ is no longer initial: there's no morphism from $\langle x, f_3, U(a) \rangle$ to $\langle x, id_x, x \rangle$ since there's no morphism $U(a) \to x$ in $\mathcal{C}$. Instead, $\langle x, id_x, x \rangle$ is initial (assuming the necessary diagrams commute).
The answer to your question, though, is yes. Universal properties can be defined using a single category (no arrow category needed). In fact, initial objects encompass all or almost all universal properties. It's all just a matter of changing which category you want the initial object to be in. The construction with the functor $U$, however, is simply a useful special case (for the category $(X \downarrow U)$) that applies to many situations. For example, limits can be defined using that construction.
It's just helpful to work out the details for special cases so you get a good general idea of what's possible. "Initial objects" be themselves don't seem very interesting, but using some special categories derived from other data (in this case, the functor $U: \mathcal{D} \to \mathcal{C}$), you get something much more interesting.
Best Answer
It's crucial to remember that fibrations over $C$ are exactly the Grothendieck constructions of contravariant (pseudo)functors from $C$ to $\mathrm{Cat},$ with the fibers of the fibration corresponding to the values of the functor. Indeed, this is really how people ought to define fibrations, with the usual definition as a characterization, since the usual definition is so hard to grasp at first.
So, what is the fiber of $\mathrm{dom}$ over some $c\in C$? It's the class of arrows with domain $c$ and commutative squares whose domain component is $\mathrm{id}_c.$ That is, the fiber is precisely the slice category $c/C.$ And we have a very simple pseudofunctor out of $C^{\mathrm{op}}$ sending $c\mapsto c/C$ and $f:c\to c'$ to the functor $c/C\to c'/C$ given by precomposition with $f.$ In fact, this happens to be a strict functor into $\mathrm{Cat}.$
For my money, that's all you should realize to understand how $\mathrm{dom}$ is a fibration. You may well find it easier to work out the proof that the Grothendieck construction of an arbitrary functor into $\mathrm{Cat}$ is a fibration than to work it out in this particular case.